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Since the ball starts from rest, we know that
T
1
=0
. Also, setting the origin of our coordinate system at
the ground, we have
0+
mg
(
H
+
h
)=
1
2
mV
2
+
mgh
Solving for
V
gives
V
=
0
2
gH
Since the initial velocity of the ball is
v
=
−
V
k
, there follows
v
t
=
t
·
v
=
V
sin
β
v
n
=
n
·
v
=
−
V
cos
β
Turning to the impact between ball and the incline, the tangential velocity component is unchanged, which
tells us that
v
I
t
=
V
sin
β
where
v
I
=
v
I
n
n
+
v
I
t
t
is the ball’s velocity just after the impact. The impact relation tells us that since the
coefficient of restitution is
e
, and since the incline is immovable,
0
−
v
I
n
=
e
(
v
n
−
0)
=
⇒
v
I
n
=
−
ev
n
=
eV
cos
β
Hence, just after the impact, the ball’s velocity is
v
I
=
eV
cos
β
n
+
V
sin
β
t
Substituting for
n
and
t
from above, we have
v
I
=
eV
cos
β
(
i
sin
β
+
k
cos
β
)+
V
sin
β
(
i
cos
β
−
k
sin
β
)
=(
eV
sin
β
cos
β
+
V
sin
β
cos
β
)
i
+
D
eV
cos
2
β
−
V
sin
2
β
i
k
Making use of the following trigonometric identities
sin
β
cos
β
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 Spring '06
 Shiflett

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